Unit 8- Solutions Aqueous Boiling point Colligative property Concentrated Dilute Molarity Parts per...
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Transcript of Unit 8- Solutions Aqueous Boiling point Colligative property Concentrated Dilute Molarity Parts per...
Unit 8- Solutions• Aqueous• Boiling point• Colligative property• Concentrated• Dilute• Molarity• Parts per million (ppm)• Percent by volume• Percent mass• Saturated• Solubility• Solute
• Solution• Solvent• Supersaturated• Unsaturated• Vapor• Vapor pressure
Mixtures
• Homogeneous– Solutions– Can be in all phases of
matter (s, l or g)• Solids- metals mixed form
an alloy. Brass= Zn + Cu• Gases- ex: air• Liquids- usually a solid or
liquid dissolved in a liquid
• Heterogeneous– Temporary mixtures are
suspensions• Ex: muddy water
– Stable mixtures are colloids• Ex: milk• Particles have charge,
they repel each other so they can’t clump and settle out
Characteristics and parts of solutions
• Characteristics– Homogeneous mixture– Are clear and don’t
disperse light– Can have color– Will not settle out– Will pass through a filter
• Parts– Solute- substance that is
being dissolved• present in a smaller
amount
– Solvent- substance that dissolves solute• Present in greater amount• Water- most common
solvent• Water solutions =
aqueous (aq)
Separating mixtures
• Physically- decanting, centrifuge, filtration, evaporation
• By solubility- chromatography
• By boiling point- distillation
Solubility • How much solute will dissolve in a certain
amount of solvent at a certain temperature– Materials with high solubility are soluble– Materials with low solubility are insoluble– “like dissolves like”– Liquids that are soluble in each other are miscible• If they aren’t they are immiscible
Factors that affect solubility• “like dissolves like”; degree of solubility (how
much dissolves)
– Soap* - long nonpolar chain with polar end• Grease= nonpolar, water= polar (soap connects the two)
– Ionic substances- dissociate- ions separate• Takes E, entropy (randomness) increases as ions
dissociate and then decreases as water molecules assemble around the ions
Solute type Non polar solvent
Polar solvent
Nonpolar Soluble InsolublePolar Insoluble SolubleIonic Insoluble soluble
Soap
PO-
CH3
CH2CH2
CH2 CH2
CH2
CH2
CH2
O-
O-
Soap
• Hydrophobic non-polar end
PO-
CH3
CH2 CH2
CH2CH2
CH2
CH2
CH2
O-
O-
Soap
• Hydrophilic polar end
PO-
CH3
CH2CH2
CH2CH2
CH2
CH2
CH2
O-
O-
PO-
CH3
CH2CH2
CH2CH2
CH2
CH2
CH2
O-
O-
_
• A drop of grease in water• Grease is non-polar• Water is polar• Soap lets you dissolve
the non-polar in the polar.
Hydrophobic ends dissolve in grease
Hydrophilic ends dissolve in water
• Water molecules can surround and dissolve grease.
Factors con’t
• Temperature– As temp of water increases, solids become more
soluble (they separate), gases become less soluble (they come together)
• Pressure– Has little or no effect on solubility in solids and liquids– Does effect the solubility of gases in liquids, as pressure
increases, solubility of a gas in the liquid increases• **greater surface area = increased dissolving
process
Factor reviewFactor Affect on Solid
Solute Affect on Gaseous
Solute
Particle Size
Stirring
Amount ofdissolved solute
Temperature
Reducing particle size by crushing increases the rate by increasing surface area.
Reducing particle size by crushing increases the rate by increasing surface area.
Not applicable
Increases the rate by exposing fresh solvent to solute and increasing kinetic energy.
Decreases the rate by increasing kinetic energy, thereby reducing solubility.
As the amount of dissolved solute increases, the rate decreases.
As the amount of dissolved solute increases, the rate decreases.
As the temperature increases, the rate increases.
As the temperature increases, the rate decreases.
Looking at solubility in the Reference Tables
• Table F - Shows ionic compounds• Predicts whether a precipitate will form when 2 ionic
compounds are mixed• A Rx will take place if 1 or both of the products are listed as
insoluble
• Table G- • Shows the amount of a
substance that can be dissolved in 100g of water at different temperatures
• Lines show the max amount of solute at a given temp
Looking at solubility in the Reference Tables
• Solids- lines that increase as temp increases
• 3 lines- NH3, HCl, SO2- show decrease in solubility as temp increases- they are GASES
• If given pt is below the line the solution is unsaturated
• If given pt is above the line the solution is supersaturated
• If given pt is on the line the solution is saturated- solubility equilibrium
TEMPERATUREM
ASS
of S
OLU
TE
UNSATURATED
SUPERSATURATED
http://www.its.caltech.edu/~atomic/snowcrystals/primer/morphologydiagram.jpg
• Saturation- occurs at a pt of solubility equilibrium
1- Add a crystal• Saturated Solution
o The crystal falls to the bottom.• Unsaturated Solution
o The crystal dissolves.• Supersaturated Solution
o Excess solute precipitates.
Saturation Test
• Which compound which is the least
soluble at 10°C?• How many grams of potassium nitrate
needed to saturate 100 mL of water at 52°C?
• One hundred mL of a potassium chloride solution is saturated at 10°C. How many additional grams are needed to saturate the solution at 50°C?
• At what temperature do saturated solutions of sodium chloride and potassium chloride contain the same mass of solute per 100 mL of water?
• How many more grams of sulfur dioxide can be dissolved in 100 mL of water at 20°C than at 90°C?
KClO3
90 g
12 g
37°C
7 g
Interpreting Solubility Curves
Determining concentration • Concentration- ratio of solute to solvent- (g/L)
• Dilute- solution with little solute• Concentrated- solution with a lot of solute
• Molarity- M- moles/liter of soln• M = moles of solute ÷ liters of solution• *many times grams must be converted to moles before
molarity is calculated• For dilute solutions- ppm- parts per million
• • •
Percent volume = Volume (Solute)
Volume (Solution)× 100%______________
Percent mass = Mass (Solute)
Mass (Solution)× 100%______________
ppm = Mass (Solute)
Mass (Solution)× 1,000,000 ppm______________
Concentration Problems
• Procedure: Divide the mass of the solute by the volume of the solvent or the solution.
What is the concentration of salt in a 20.0 mL solution containing 18.3 g of dissolved NaCl?
Concentration = 18.3 g20.0 mL = 0.92 g/mL__________
• Step 1: Find the mass of the solution10.0 g + 2.3 g = 12.3 g
• Step 2: Divide the mass of the solute by the mass of the solution and multiply by 100 %
Percent by Mass ProblemsWhat is the percent by mass of ethanol in a solution containing 2.3 g of ethanol (C2H5OH) dissolved in 10.0 g of water?
Percent mass = 2.3 g12.3 g × 100% = 19 %__________
Percent by Volume Problems
• Step 1: Find the volume of the solution18.2 mL + 85.0 mL = 103.2 mL
• Step 2: Divide the volume of the solute by the volume of the solution and multiply by 100 %
What is the percent by volume of glycerine in a solution containing 18.2 mL of glycerine (C3H6O3) dissolved in 85.0 mL of water?
Percent volume = 18.2 mL103.2 mL × 100% = 17.6 %______________
Parts per Million Problems
• Step 1: Find the mass of the solution10.0 g + 0.0035 g = 10.0035 g
• Step 2: Divide the mass of the solute by the mass of the solution and multiply by 1,000,000 ppm.
About 0.0035 g of hydrogen sulfide are dissolved in10.0 g of water. Express this in parts per million.
ppm = 0.0035 g10.0035 g × 1,000,000 ppm = 350 ppm______________
• Step 1: Convert all volumes to liters
• Step 2: Substitute values into the definitional equation
Find the molarity of 100. mL of a solution that contains 0.25 moles of dissolved solute.
100. mL × 0.001 L____________
1 mL= 0.100 L
M = 0.25 mol____________
0.100 L= 2.5 M
Molarity Sample Problems
• Step 1: Find the GFMNa = 23 × 1 = 23O = 16 × 1 = 16H = 1 × 1 = 1
40• Step 2: Convert all volumes to liters
• Step 3: Substitute values into the correct equation
Find the molarity of 250. mL of a solution that contains 4.0 g of dissolved sodium hydroxide (NaOH).
Sample Problem 2
250. mL × 0.001 L____________
1 mL= 0.250 L
• Step 1: Convert all volumes to liters
• Step 2: Substitute values into the correct equation
How many moles of solute are dissolved in 30 mL of a2 M solution?
Sample Problem 3
30 mL × 0.001 L____________
1 mL= 0.03 L
0.06 mol
• Step 1: Find the GFMAg = 108 × 1 = 108N = 14 × 1 = 14O = 16 × 3 = 48
170• Step 2: Convert all volumes to liters
• Step 3: Substitute values into the correct equation
How many grams of silver nitrate (AgNO3) are needed to prepare 200. mL of a 0.10 M solution?
Sample Problem 4
200. mL × 0.001 L____________
1 mL= 0.200 L
= 3.4 g
How to prepare a solution• Add desired amount of solute to volumetric flask• Add some DI water and mix until dissolved• Fill flask to the line on the neck, close and mix • What mass of sodium bicarbonate is required to
prepare 2 L of a .250M sodium bicarbonate solution?– Knowns: concentration= .250M vol of soln= 2L - unknown: mass of Na2CO3 = ____g
- Find moles of using molarity equation then convert moles to grams
Preparing Solutions sample problems:
• How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution?
• How many grams of CaCl2 are needed to make 625 mL of a 2.0 M solution?
Physical Properties of Solutions
• Conductivity-– Depends on solution having charged particles that
can move– Electrolyte- substances that conduct electricity
when melted or dissolved in water (ionic substances)– Sports drinks etc.
• Strong electrolytes- completely dissociate• Weak electrolytes- only some ions dissociate (weak
conductivity)
– Nonelectrolyte- no ions; doesn’t conduct (molecular substances)
Colligative Properties
• Dependent on # of dissolved particles– A physical effect of solute on solvent– Degree of effect determined by conc. of dissolved
particles
• Freezing pt depression, boiling pt elevation, lowering vapor pressure
Freezing Point depression
• Crystallization process in water is disrupted by solute particles
• Salt on sidewalks and roads• Making ice-cream• 1 mole of any particles will have the same
effect on freezing pt of water– 1 mole of particles lowers freezing pt 1.86oC of
1000g of water
Vapor Pressure
• The pressure that a vapor exerts
• Table H shows vapor pressure of liquids at different temperatures
BP elevation and VP lowering
• Boiling pt- when VP = atmospheric pressure– As temp increases, VP increases until it reaches
atmospheric pressure; then it starts to boil• Table H will show at what temp water will boil
at different pressures• As solute is added to a solution, vapor
pressure decreases; therefore it takes longer for the VP to reach the atmospheric pressure and the water boils at a higher temperature